Bài 1:
a) $x^3 - x^2 - x - 2$
$= x^3 - 2x^2 + x^2 - 2x + x - 2$
$= x^2(x- 2) + x(x - 2) + x - 2$
$= (x -2)(x^2 + x + 1)$
b) $x^2 + 2xy + y^2 - x - y - 12$
$= (x+y)^2 - (x + y) - 12$
Đặt $t = x + y$ ta được:
$t^2 - t - 12$
$= t^2 - 4t + 3t - 12$
$= t(t -4) + 3(t - 4)$
$= (t - 4)(t + 3)$
$= (x + y - 4)(x + y + 3)$
c) $(x^2 + x + 1)(x^2 + x + 2) - 12$
Đặt $t = x^2 + x + 1$ ta được:
$t(t + 1) - 12$
$= t^2 + t - 12$
$= t^2 + 4t - 3t - 12$
$= (t+4)(t - 3)$
$= (x^2 + x + 5)(x^2 + x - 2)$
Bài 2:
$P = 4x^2 - 6xy + 9y^2- 16x + 12y + 2020$
$= \dfrac{1}{4}(16x^2 + 9y^2 + 64 - 24xy - 64x + 48y) + \dfrac{27}{4}y^2 + 2004$
$= \dfrac{1}{4}(4x -3y - 8)^2 +\dfrac{27}{4}y^2 + 2004$
Ta có:
$\begin{cases}(4x - 3y - 8)^2 \geq 0, \, \forall x,y\\y^2 \geq 0, \, \forall y\end{cases}$
Do đó:
$\dfrac{1}{4}(4x -3y - 8)^2 +\dfrac{27}{4}y^2 + 2004 \geq 2004$
Dấu = xảy ra $\Leftrightarrow \begin{cases}4x - 3y - 8 = 0\\y = 0\end{cases}\Leftrightarrow \begin{cases}x = \\y = 0\end{cases}$
Vậy $\min P = 2004 \Leftrightarrow (x;y) = (2;0)$
